find the principal unit normal vector to the curve
find the principal unit normal vector to the curve at the specific value of the parameter.
 = ti + \frac {1}{2}t^2j, t = 2)
i need to use this formula
 = \frac {r'(t)}{ || r'(t) ||} )
so, to find r'(t)
 = i + t j )
so, now to find the magnitude: simplified
 || = \sqrt {1 + t^2} )
puting them together
 = \frac {i + t j} {\sqrt {1 + t^2}} )
From here is where i get lost
I need to find T'(t)
I can seem to follow the solutions manual. I think they are using the quiotient rule due to the denominator having the 3/2
fraction. But if they are i still cant see how they are doing it.
Can anyone help me work out the next step for T'(t)?
 = \frac { -t}{(t^2 + 1)^{3/2}}i + \frac {1}{ (t^2 + 1) ^{3/2}} j)
Thank you so much!!!
Re: find the principal unit normal vector to the curve
 = \frac{\mathbf{i} + t\mathbf{j}}{\sqrt{1 + t^2}})
 = \left(1+t^2\right)^{-1/2}\mathbf{i} + t\left(1+t^2\right)^{-1/2}\mathbf{j})
Differentiating,
![\mathbf{T}'(t) = -\frac12(2t)\left(1+t^2\right)^{-3/2}\mathbf{i} + \left[-\frac12t(2t)\left(1+t^2\right)^{-3/2} + \left(1+t^2\right)^{-1/2}\right]\mathbf{j}](http://latex.codecogs.com/png.latex?\mathbf{T}'(t) = -\frac12(2t)\left(1+t^2\right)^{-3/2}\mathbf{i} + \left[-\frac12t(2t)\left(1+t^2\right)^{-3/2} + \left(1+t^2\right)^{-1/2}\right]\mathbf{j})
![\Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \left[\frac{-t^2}{\left(1+t^2\right)^{3/2}} + \frac1{\left(1+t^2\right)^{1/2}}\right]\mathbf{j}](http://latex.codecogs.com/png.latex?\Rightarrow\mathbf{T}'(t) = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \left[\frac{-t^2}{\left(1+t^2\right)^{3/2}} + \frac1{\left(1+t^2\right)^{1/2}}\right]\mathbf{j})
 = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac{-t^2+\left(1+t^2\right)}{\left(1+t^2\right)^{3/2}}\mathbf{j})
 = \frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j})
We find the magnitude,
\right\| = \sqrt{\frac{t^2+1}{\left(1+t^2\right)^3})
\right\| = \sqrt{\frac1{\left(t^2+1\right)^2})
\right\| = \frac1{t^2+1})
And therefore,
 = \frac{\mathbf{T}'(t)}{\left\|\mathbf{T}'(t)\right \|})
![=\left(t^2+1\right)\left[\frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}\right]](http://latex.codecogs.com/png.latex?=\left(t^2+1\right)\left[\frac{-t}{\left(1+t^2\right)^{3/2}}\mathbf{i} + \frac1{\left(1+t^2\right)^{3/2}}\mathbf{j}\right])
![=\frac{-t}{\sqrt{1+t^2}}\mathbf{i} + \frac1{\sqrt{1+t^2}}\mathbf{j}\right]](http://latex.codecogs.com/png.latex?=\frac{-t}{\sqrt{1+t^2}}\mathbf{i} + \frac1{\sqrt{1+t^2}}\mathbf{j}\right])
Re: find the principal unit normal vector to the curve
in your second step the first arrow before Differentiating, im not sure how you got j
for i did you just bring up the denominator? doesnt make sense
Re: find the principal unit normal vector to the curve
Quote:
Originally Posted by
icelated 
in your second step the first arrow before Differentiating, im not sure how you got j
for i did you just bring up the denominator? doesnt make sense
Did you forget these properties of exponents from algebra?
![x^{m/n} = \sqrt[n]{x^m}](http://latex.codecogs.com/png.latex?x^{m/n} = \sqrt[n]{x^m})
Re: find the principal unit normal vector to the curve
Wow, okay how did i miss that! I wasnt splitting the i and j up. Once i did now i can see the answer fall out.
